Find Intervals of Increase and Decrease for y = (4x + 22)²

To solve this problem, we'll begin by finding the derivative of the given function $y = (4x + 22)^2$ with respect to $x$.

The function can be expanded as:

$y = (4x + 22)^2 = 16x^2 + 2 \cdot 4 \cdot 22 \cdot x + 22^2 = 16x^2 + 176x + 484$

Next, find $\frac{dy}{dx}$ by differentiating:

$\frac{dy}{dx} = \frac{d}{dx}(16x^2 + 176x + 484) = 32x + 176$

Set the derivative equal to zero to find the critical point:

$32x + 176 = 0$

$32x = -176$

$x = -\frac{176}{32}$

$x = -\frac{11}{2}$

$x = -5.5$

This critical point, $x = -5.5$, will be the vertex of the parabola, determining where the function changes from decreasing to increasing.

Now, test the sign of $\frac{dy}{dx} = 32x + 176$ in intervals around the critical point:

• For $x < -5.5$, choose $x = -6$:

$32(-6) + 176 = -192 + 176 = -16$

Since the derivative is negative, the function is decreasing.

• For $x > -5.5$, choose $x = -5$:

$32(-5) + 176 = -160 + 176 = 16$

Since the derivative is positive, the function is increasing.

Therefore, the intervals of increase and decrease are:

$\searrow:x < -5\frac{1}{2}$

$\nearrow:x > -5\frac{1}{2}$

Thus, the correct choice from the given options is:

$\searrow:x < -5\frac{1}{2}$

$\nearrow:x > -5\frac{1}{2}$